Question

From experience, it is known that the life expectancy of tires is normally distributed. An independent agency tests a random sample of 6 tires and finds the following life expectancies (in thousands of miles): 26, 34, 28, 31, 22, and 30. a) Find the sample mean =

b) Find the confidence interval and the margin of error, E, (Round to 2 decimal points) for the population mean life expectancy of tires for 98% confidence level. Use s = 4.18. 98% confidence interval = _______________ E = ____________________

c) Suppose a manufacturer claims that the mean life expectancy for tires is 33,000 miles. Does the confidence interval above support this claim? ______________________

Answer 1

Part a

Sample mean = 28.5

Part b

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 28.5

S = 4.18

n = 6

df = n – 1 = 5

Confidence level = 98%

Critical t value = 3.3649

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 28.5 ± 3.3649*4.18/sqrt(6)

Confidence interval = 28.5 ± 5.7422

Lower limit = 28.5 - 5.7422 = 22.76

Upper limit = 28.5 + 5.7422 = 34.24

Confidence interval = (22.76, 34.24)

Margin of error = t*S/sqrt(n) = 3.3649*4.18/sqrt(6) = 5.7422

Margin of error = 5.74

Part c

Yes, this confidence interval supports this claim because the value 33,000 is lies between above interval.